Day 3 - Sum Pair closest to target.md

Difficulty	Source	Tags
Easy	160 Days of Problem Solving	two-pointer-algorithm Arrays

🚀 Day 3. Sum Pair Closest to Target 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given an array arr[] and a number target. Your task is to find a pair of elements (a, b) in arr[] where a <= b whose sum is closest to the target. If there are multiple such pairs, return the pair with the maximum absolute difference. If no such pair exists, return an empty array.

🔍 Example Walkthrough:

Input:
arr[] = [10, 30, 20, 5], target = 25
Output:
[5, 20]

Input:
arr[] = [5, 2, 7, 1, 4], target = 10
Output:
[2, 7]
Explanation: As both (4, 7) and (2, 7) are closest to the target 10, the absolute difference of (2, 7) is 5 and (4, 7) is 3. Hence, [2, 7] is the correct pair.

Input:
arr[] = [10], target = 10
Output:
[]
Explanation: The array contains only one element, so no valid pair can be formed.

Constraints:

• $1 <= arr.size() <= 2*10^5$
• $0 <= target <= 2*10^5$
• $0 <= arr[i] <= 10^5$

🎯 My Approach:

1. Sorting and Two-pointer Approach:
   • First, sort the array.
   • Use two pointers, l (left) and r (right), to traverse the array. Start by initializing l at the beginning and r at the end.
   • Calculate the sum of arr[l] and arr[r] and check how close it is to the target. Track the pair that gives the smallest difference.
   • Move the pointers based on whether the sum is less than or greater than the target, to attempt getting closer to the target.
   • If multiple pairs have the same sum, the one with the largest absolute difference should be returned.
   • Continue this process until the pointers meet.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n log n), where n is the number of elements in the array. This is because we need to sort the array, and sorting takes O(n log n) time. After sorting, the two-pointer traversal takes O(n) time.
• Expected Auxiliary Space Complexity: O(1), as we only use a constant amount of additional space (for the two pointers and result storage).

📝 Solution Code

Code (C++)

class Solution {
public:
    vector<int> sumClosest(vector<int>& arr, int target) {
        sort(arr.begin(), arr.end());
        vector<int> res;
        for (int l = 0, r = arr.size() - 1, minDiff = INT_MAX; l < r;) {
            int sum = arr[l] + arr[r], diff = abs(target - sum);
            if (diff < minDiff) minDiff = diff, res = {arr[l], arr[r]};
            sum < target ? ++l : --r;
        }
        return res;
    }
};

Code (Java)

class Solution {
    public List<Integer> sumClosest(int[] arr, int target) {
        Arrays.sort(arr);
        int l = 0, r = arr.length - 1, minDiff = Integer.MAX_VALUE;
        List<Integer> res = new ArrayList<>();
        while (l < r) {
            int sum = arr[l] + arr[r], diff = Math.abs(target - sum);
            if (diff < minDiff) {
                minDiff = diff;
                res = Arrays.asList(arr[l], arr[r]);
            }
            if (sum < target) l++;
            else r--;
        }
        return res;
    }
}

Code (Python)

class Solution:
    def sumClosest(self, arr, target):
        arr.sort()
        l, r, minDiff, res = 0, len(arr) - 1, float('inf'), []
        while l < r:
            s = arr[l] + arr[r]
            if abs(target - s) < minDiff:
                minDiff, res = abs(target - s), [arr[l], arr[r]]
            l += s < target
            r -= s >= target
        return res

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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